question_answer

A body is moving from A which is at a height of 25 m from the ground and follows the path as shown in the figure. Find the velocity of the body on reaching 'B' which is at a height of 5 m from the ground.

Answer:

According to law of conservation of Energy, Total Energy at A = Total Energy at B \[mg{{h}_{1}}+\frac{1}{2}mv_{1}^{2}=mg{{h}_{2}}+\frac{1}{2}mv_{2}^{2}\] But velocity at highest point \[A={{v}_{1}}=0\] \[\therefore \,\,mg{{h}_{1}}=mg{{h}_{1}}+\frac{1}{2}mv_{2}^{2}\] \[\Rightarrow \frac{1}{2}mv_{2}^{2}=mg{{h}_{1}}-mg{{h}_{2}}=mgh\,\,({{h}_{1}}-{{h}_{2}})\] \[\therefore V_{2}^{2}=\frac{2mg\,\,({{h}_{1}}-{{h}_{2}})}{m}\] \[=2\times 10(25-5)=20\times 20=400\] \[\therefore \,{{v}_{2}}=20\,\,m/s\]